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CFMs: Duct Size vs. Heating Load

[email protected] | Posted in General Questions on

HVAC question.

Building a house in zone 6 next year (Cumberland County Me)… not to passive standards but pretty tight.  Zip R sheathing…closed cell and minreal wool for insulation…r30 walls, r50 ceiling.  1700ft2 upstairs and 1250 walkout basement downstairs… using some manual J calculators recommended in other GBA discussions I come up with about 45000btu/hr heating requirement for the house.

Equipment I am planning using is a 48k Mitsubishi ducted heat pump with hyper heat and a ERV for ventilation.

Finally to my question… when I am figuring out the duct work size for a master bed bedroom… 200ft2 x 9ft ceiling = 1800ft3 x 5 (recommended ACH) = 9000ft3/hr / 60 = 150cfm

but heating room load is 4000btu/hr for the room
cfm = 4000btu/hr divided by (1.08 x 72F) = 53 cfm… this seems low to me.

which one is the driver?  When I see 2500ft2 passive houses heated by a 1.5 to 2 ton unit there is no way they are getting the recommended ACH… do I just disregard the recommended ACH since I am bringing in 150cfm to the house thru the ERV and concentrate on heat loads?

thanks for any info,
Steve

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Replies

  1. Expert Member
    NICK KEENAN | | #1

    Where are you getting the 72F? That should be the temperature difference between the incoming air and the room air. That's probably closer to 30F which gives you 133 CFM.

    1. [email protected] | | #2

      I figured it was the winter 99% design temp for that area of -2f to the desired room temp of the room at 70f. Not the case I guess? But like I said in the post 53cfm seemed pretty low.

      1. Expert Member
        Akos | | #3

        The temperature increase in a typical heat pump is much less than a gas furnace. You typically don't see a 72F increase in the air stream, usually it is closer to 30F delta. This does give higher flow rate for the room.

        But you do touch on the issue with a centrally ducted ERV system. You need to rely on the HVAC to provide sufficient airflow to satisfy both heating/cooling and ventilation needs. This becomes an issue for low load homes as they need much less heat, thus CFM, in each room. This is why the general consensus is the best system is a fully ducted ERV where the ventilation flow is decoupled from heating/cooling system.

        I would also re-run your heat load calculations for the place, for something with R30 walls and R50 roof, 48000 BTU is excessively high. I would guess the real heat load is closer to 20000BTU. Lot of online calculators have very high default air leakage rates. With a bit of care on a new build, it should be easy to be sub 2ACH50.

        P.S. SPF is something you generally design out of new builds. There are much better, cheaper and way more environmentally friendly ways to air seal and insulate a new build.

        1. [email protected] | | #4

          Thanks,
          Thanks for the words... as I get closer to the build I am planning on paying for some another manual j calculation program to get some better numbers. I am planning on a separated ducted ERV.
          As for the insulation I am planning on air sealing with knauf ecoseal or Dap airstop... but I am still debating if I’m going to do a flash and batt with closed cell and mineral wool or go thicker on the zip r and r23 mineral wool to get the r30 walls.

          1. Expert Member
            Akos | | #6

            Flash and batt doesn't significantly increase the assembly R value because of the thermal bridging of the studs. Flash and bat makes the most sense when there is no easy way to air seal such as older houses with lumber sheathing.

            For new build, the simplest is to tape your sheathing. Air sealing with caulking also works, I find it takes more work and even than it is hard to get a good joint.

            For higher R value assembly you can read through this:

            https://vancouver.ca/files/cov/ig-r22-effective-walls-residential-construction.pdf

            I used to say that a cheap high R value wall is a 2x8 with HD batts and rain screen (no exterior insulation), but with the high cost of lumber, I think a 2x4 wall with thick exterior insulation is the way to go.

          2. Expert Member
            Dana Dorsett | | #8

            What Akos said.

            BTU, 45K for a 3000' house is pretty high- many code-min houses can hit that with a reasonably careful design. I suspect this is an issue of a relative newbie compounded by less-flexible freebie online tools that tend to overshoot reality by 25-30% or more even with reasonably aggressive inputs.

            How much of that 45K load number is ventilation + infiltration?

            >"I am still debating if I’m going to do a flash and batt with closed cell and mineral wool or go thicker on the zip r and r23 mineral wool to get the r30 walls."

            Closed cell foam between studs is a waste (end of story full stop) due to the thermally bridging framing. Do the math:

            https://www.finehomebuilding.com/membership/pdf/184243/021269086NRGnerd.pdf

            From an energy point of view the whole-assembly R value (with the thermal bridging factored in) is far more important than the center-cavity R. Going with fatter ZIP-R delivers more performance per additional-R, but in zone 6 the ZIP needs to be more than ~35% of the total center-cavity R for reasonable dew point control at the foam/fiber boundary. With R23 batts that means only the 2.5" thick R12.6 ZIP-R will do. ( R12.6/[R12.6 + R23] = ~35%).

            At 2.5" thickness the ZIP-R isn't sufficiently structural against racking forces no matter how tight the fasterr spacing, so flat metal or let-in X-bracing or shear panels would need to be designed in to keep the bad nor'easter wolves from huffing & puffing and blowing the house down some day.

      2. charlie_sullivan | | #5

        The temperature that goes in that calculation is the temperature difference between the room air and the supply air from the heating system: 100-70, for example. The 99% design temperature is used in finding the 4000 BTU/h.

        1. [email protected] | | #7

          Thanks for the clarification, make more sense.

  2. HFF | | #9

    All the comments here raise good points. To directly adress your question, here is how to calculate cfm for your heat. From the manufacturer, obtain the maximum fan output of your heat pump unit. Then divide that flow up according to your room to room heat losses. In your example, total heat loss is given as 45000 BTU/h, and your bedroom loses 4000. So 4000/45000 = 9% of your total fan's output goes to that room. Of course then you have to design the ductwork to deliver that amount. Everything should add up to your max fan cfm. Owing to duct losses, you'll never get that total flow, but you have apportioned it properly.

  3. HFF | | #10

    PS your ERV ventilation rate is way off. 150 cfm is about what an entire house needs, not a 200 sq ft bedroom! 5 ACH is more in line with the needs of a medical laboratory.

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